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Math Help - nth moments using moment generating functions

  1. #1
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    nth moments using moment generating functions

    For a standard random variable Z let \mu_n = E[Z^n]
    Show that:

    \mu_n = 0 when n is odd
    \mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j

    I started:
    E[e^tZ]=e^{{t^{2}}/2} from moment generating function of a normal variable
    = \displaystyle\sum_{j=0}^\infty \frac{(t^2/2)^j}{j!}
    = \displaystyle\sum_{j=0}^\infty \frac{t^{2j}}{2^jj!}


    however I cannot see where to go now
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  2. #2
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    Quote Originally Posted by FGT12 View Post
    For a standard random variable Z let \mu_n = E[Z^n]
    Show that:

    \mu_n = 0 when n is odd
    \mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j

    I started:
    E[e^tZ]=e^{{t^{2}}/2} from moment generating function of a normal variable
    = \displaystyle\sum_{j=0}^\infty \frac{(t^2/2)^j}{j!}
    = \displaystyle\sum_{j=0}^\infty \frac{t^{2j}}{2^jj!}


    however I cannot see where to go now
    I assume you mean a standard normal random variable then

    \mu_n=\mathbb{E}[z^n]=\frac{1}{2\pi \sigma^2}\int_{-\infty}^{\infty}z^ne^{\frac{-(z-\mu)^2}{2\sigma^2}}

    The Gaussian is an even function and if  \mu=0 then

    z^n

    Is even or odd depending on the exponent.

    So what do you know about integrals about even and odd functions over symmetric intervals?
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  3. #3
    MHF Contributor matheagle's Avatar
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    what is j in this???


    \mu_n = 0 when n is odd
    \mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j

    Ok the math killed the english, I see it when I quote you.

    \mu_n = \frac{(2j)!}{2^jj!} when n=2j

    why not just write

    \mu_n = \frac{n!}{2^{n/2}(n/2)!} if n is even

    Did you try differentiating the MGF?
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  4. #4
    MHF Contributor matheagle's Avatar
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    I would just use the gamma function

    E(Z^n)={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}z^ne^{-z^2/2}dz

    =\sqrt{2\over \pi}\int_0^{\infty}z^ne^{-z^2/2}dz

    Then let w=z^2/2
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